last minute help!

[Jo07]

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I'm taking the DAT tomorrow and have two areas that I am having problems with:

HNMR and aromaticity--I can't weed out the one 'trick' answer that always seems to be there. Don't get me wrong, I have studied both of these subjects but they are still giving me problems! If anyone has any pointers or tips on how to quickly and efficiently weed out the wrong answers I would appreciate it.

 

[DanNuch]

can you be a bit more specific with the aspect that is giving you some trouble?

 

[Jo07]

well mostly the HNMR....

I am having a hard time telling how many peaks there are. for each 'set' of equivalent H's you do n+1 for the peak, but I can't tell which ones are equivalent!

 

[Orgodox]

Hey Jo07 I hope I can help you out on the aromatic issue. Many people find this confusing but its so overrated and not that bad...
Lets start with the 4 rules. Im sure you have heard the 4 rules many times but bear with me I think this will help.

Cyclic
Fully conjugated
Huckles rule (4n+2)
Planar

So right away when you see something and want to know if it might be aromatic check to see if its a ring. If it is not then you know it 100% is not aromatic if it is then continue on to number 2

2. Fully conjugated-- If you see that one of the carbons in the ring is sp3 hybridized then you know its not aromatic. For example you have a carbon with no double bonds and 2 Hydrogens it cant be aromatic because the electrons cant delocalize ( they cant move around because that sp3 carbon cant help spread around the charge. If however there is a positive charge at a carbon then that means there is a vacant p orbital that can help with delocalization, for example the cycloheptatrienyl cation, and you should continue to 3 to see if it can be aromatic...)

3. Huckles rule: So the number of PI electrons needs to satisfy the 4n+2 rule where n = an integer. n can be 0,1,2,3... so for example in benzene there are 6 pi electrons so 4n+2=6 and n = 1 so that would pass the test. (As an aside note if there are 4 or 8 or 12 pi electrons IE 4N pi electrons then this is destabilizing and called anti-aromatic)

So here comes the part where I think many people are confused. What if there is a heteroatom (non carbon or hydrogen atrom) in the ring and it has lone pairs, do the lone pairs contribute tot he pi electrons or not? (We can also ask this by a carbon with a negative charge for that matter as in cyclopentadienyl anion)
You have to realize that the atom with the lone pair in question can only have ONE pair of electrons in the Pi system and would be counted towards the pi electrons to apply the 4n+2 rule. So if the atom is already doubly bonded to a neighboring atom then you know its lone pair cant be in the PI system. Additionally if the atom has two lone pairs only one of them will be in the pi system. For example Furan. Only of the lone pairs should be counted but the other CANT because its electrons are not in the same plane.

Just to be 100% clear lets take a hard example Imidazole

Looking at the form all the way ont he left. The nitrogen with the hydrogen has a lone pair. It does not have any double bonds associated with it in this resonance form and the lone pair can be included in the PI system. The other Nitrogen though already has a double bond. Its lone pair must lie in a different plane and is not part of the pi system and should not be counted. Since we have 6 Pi electrons this is aromatic


4. About planarity You don't really need to know much if it passes these 3 criteria I think its safe to assume for the DAT that its aromatic (even though theres more to it)

I hope this helps clarify thing. Good luck on your test!

 

[Willem]

Good info there!


Took me a while to learn all these details myself as my teacher kind of skimmed over the whole "aromatic or not" section. Wish Id had something as concise as this.

 

[Jo07]

thanks so much!! Very helpful! Hmm..perhaps someone can help me out with this HNMR?

What is the process behind finding splitting pattern of something like:


CH3-CH2-CH2-CH=0

In the destroyer it specifies for the gamma C, but in general how would I go about attacking this problem??

 

[Jo07]

number 158 in destroyer...


why does ans b give a singlet? shouldn't it give a triplet because the next C with H on it is the CH2 group so (n+1)?

So confused 😕

 

[Orgodox]

thanks so much!! Very helpful! Hmm..perhaps someone can help me out with this HNMR?

What is the process behind finding splitting pattern of something like:


CH3-CH2-CH2-CH=0

In the destroyer it specifies for the gamma C, but in general how would I go about attacking this problem??

Hey glad the aromatics went well. Ill give the HNMR shot now...

To determine the splitting pattern check how man hydrogens the neighboring carbon has...If the neighboring carbon has 2 hydrogens then the splitting pattern will be n+1 =3 ie triplet and if it has has 3 carbons then it will be 3+1= quartet.

So for this example lets start with the alpha hydrogen. We go to the carbon bearing these Hydrogens and move to its neighbor. The neighbor is the beta carbon which has 2 Hydrogens so the alpha hydrogens would give a splitting pattern of 2+1= triplet.

Now for the betta Hs-- check the neighboring carbonS (this one is tricky you need to check both ways) to the right the alpha carbon has 2 hydrogens and to th left the gamma carbon has 3 so we have a triples of quartets.

You said the gamma Hs were done by Dr. Ramano and he can definitely explain things better than me so ill leave that to the destroyer for the answer.

 

[Orgodox]

number 158 in destroyer...


why does ans b give a singlet? shouldn't it give a triplet because the next C with H on it is the CH2 group so (n+1)?

So confused 😕

So the most downfield one here is the carbon directly attached to the O (the CH2 group). Since its neighbor has 3 Hs (see above or even better see destroyer for explanation) it gives a quartet which is the most downfield in this case right around 4.

Next downfield is the singlet you refer to. This is attached to the carbonyl carbon. This is not the splitting pattern for the CH2 group as you said the CH2 group should be a quartet. This terminal methyl give a singlet and is second most downfield.
Just for completeness the other CH3 group is most upfield. Its neighboring Carbon (The aplha carbon) has 2Hs and 2+1=3 so splits triplet which is exactly what you see in the graph!

Does that answer the question? Hope it helps!

 

[Jo07]

so if a C doesn't have any other C with H on them directly attached, it will always be a singlet?

 

[Jo07]

I think I understand this now! Thank you so much for taking the time to explain this. It's much appreciated!👍

 

[DanNuch]

yea, in a simplified way of thinking, you only look at the adjacent Carbons to figure if it is a singlet, doublet, triplet, etc.

like in 158, the Carbon on the far left is next to 0 carbons with H's, so (0+1) it is a singlet... the CH2 carbon is next to a CH3 so they represent a (1+3) quartet and (2+1) triplet respectively...

hope this helps, its hard to explain purely on words and no pictures for me...